Is this limit correct: $\lim_{x \to+\infty} \frac{\log_{2}(x-1)}{x} = 0$?

Using change of base for logarithms, you can write $$\log_2(x-1)=\frac{\ln(x-2)}{\ln(2)}$$ so we have $$\underset{x\to\infty}{\lim}\frac{\ln(x-1)}{x\ln(2)}$$ Notice as $x\to\infty$, we get "$\frac{\infty}{\infty}$" and so we can use L'Hospital's rule and take derivatives of the numerator and denominator to get $$\underset{x\to\infty}{\lim}\dfrac{\frac{1}{x-1}}{\ln(2)}=\underset{x\to\infty}{\lim}\dfrac{1}{\ln(2)(x-1)}=0$$